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Considering the alarming importance of CSAT in UPSC CSE Prelims exam and with enormous requests we received recently, InsightsIAS has started Daily CSAT Test to ensure students practice CSAT Questions on a daily basis. Regular Practice would help one overcome the fear of CSAT too.
We are naming this initiative as Insta– DART – Daily Aptitude and Reasoning Test. We hope you will be able to use DART to hit bull’s eye in CSAT paper and comfortably score 100+ even in the most difficult question paper that UPSC can give you in CSP-2021. Your peace of mind after every step of this exam is very important for us.
Looking forward to your enthusiastic participation (both in sending us questions and solving them on daily basis on this portal).
Wish you all the best ! 🙂
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Question 1 of 5
1. Question
An 8 – digit number 4952746B leaves remainder when divided by 3. How many values of b are possible?
Correct
Sol (c)
Given number = 4252746B
Sum of digits = 4 + 2 +5+ 2 + 7 + 4 + 6 + B = 30 + B
We know that a number whose sum of digits is divisible by 3 is also divisible by 3. Hence, the possible values of B can be 0 , 3 , 6 of 9 . So , there are four possible values of B.
Incorrect
Sol (c)
Given number = 4252746B
Sum of digits = 4 + 2 +5+ 2 + 7 + 4 + 6 + B = 30 + B
We know that a number whose sum of digits is divisible by 3 is also divisible by 3. Hence, the possible values of B can be 0 , 3 , 6 of 9 . So , there are four possible values of B.
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Question 2 of 5
2. Question
If the numerator and denominator of a proper fraction are increased by the same positive quantity which is greater than zero, the resulting fraction
Correct
Sol. (b)
Let the fraction be 1/ 2
Suppose we add a number 3 to both numerator and denominator, we get
1 +3/2 + 3 = 3/5
Now 1/2, = 0.5
And 3/5 = 0.6
3/5 > 1/2
Hence, the new fraction is always greater than the original fraction.
Incorrect
Sol. (b)
Let the fraction be 1/ 2
Suppose we add a number 3 to both numerator and denominator, we get
1 +3/2 + 3 = 3/5
Now 1/2, = 0.5
And 3/5 = 0.6
3/5 > 1/2
Hence, the new fraction is always greater than the original fraction.
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Question 3 of 5
3. Question
A printer numbers the pages of a book starting with 1 and uses 3089 digits in all . How many pages does the book have?
Correct
Sol. (c)
Page number 1-9
10-99
100-999
Number of digit
1x 9 = 9
2 x 90 = 180
3×900 = 2700
Total = 2889
Hence, digits left 3089 – 2889 = 200.
Now from page number 1000 to above, we have 4 digits on each page .
So. Number of pages covered in 20 digital
= 200/4
= 50
So required total number of pages
Incorrect
Sol. (c)
Page number 1-9
10-99
100-999
Number of digit
1x 9 = 9
2 x 90 = 180
3×900 = 2700
Total = 2889
Hence, digits left 3089 – 2889 = 200.
Now from page number 1000 to above, we have 4 digits on each page .
So. Number of pages covered in 20 digital
= 200/4
= 50
So required total number of pages
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Question 4 of 5
4. Question
Consider two statements SI and S2 followed by a question
S1: p and q both are prime numbers.
S2: p + q is an odd integer,
Question: Is pq an odd integer?
Which one of the following is correct?
Correct
Sol. (d)
S1: p and 9 both are prime numbers, this information is necessary.
S2: p + q is an odd integer because, the sum of even and odd always make odd. To answer the question, we must have both the statements.
Hence, both S1 and S2 are together required to answer the question.
Incorrect
Sol. (d)
S1: p and 9 both are prime numbers, this information is necessary.
S2: p + q is an odd integer because, the sum of even and odd always make odd. To answer the question, we must have both the statements.
Hence, both S1 and S2 are together required to answer the question.
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Question 5 of 5
5. Question
Number 136 is added to 5B7 and the sum obtained is 7A3, where A and B are integers. It is given that 7A3 is exactly divisible by 3. The only possible value of B is
Correct
Sol. (d)
Now from hit and trial method if 7A3 have to be divisible by 3 then 7 + A + 3 = 10 + A. The minimum value which A should have is 2 so that the sum of digits become 12 , which is exactly divisible by 3 . Now, the number is 723.
5B7 = 7A3 – 136
5B7 = 723 – 136
5B7 = 587
By, comparing both sides, we have B = 8.
Incorrect
Sol. (d)
Now from hit and trial method if 7A3 have to be divisible by 3 then 7 + A + 3 = 10 + A. The minimum value which A should have is 2 so that the sum of digits become 12 , which is exactly divisible by 3 . Now, the number is 723.
5B7 = 7A3 – 136
5B7 = 723 – 136
5B7 = 587
By, comparing both sides, we have B = 8.
